Detangling the Math

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HarpersbazarYou know how some math problems start off simple and then, the more you look at them, the more complex they appear, with hidden depths and meanings?

This isn’t one of those.  This one needs a sign with “No diving allowed.”  Nonetheless, it was surprising to me in its shallowness (like the joke about the smallest not-interesting whole number:  hey, isn’t that interesting?).

I wore my hair down yesterday, but I had to braid it at night because I knew it it would get tangled up something fierce if I slept with it down.  That got me wondering about the mathematical reason for the number of tangles.  In particular, how does the number of possible tangles decrease as you partition hair into three segments for braiding, assuming that tangles could still occur within a segment but not between the three segments?

Now with braids, a big reason that there isn’t much tangling is because of the added tension on the strands of hair.  So let’s switch to something else instead:  spaghetti.

Suppose you put n strands of spaghetti into 3 pots, and counted the possible pairs that could tangle.  There would be \binom{n}{2} pairs in each pot, for a total of 3 \cdot\binom{n}{2} pairs overall.  On the other hand, if you’d dumped all 3n pieces into a single pot, there would be {3n}\choose{2} pairs.

If you look at the ratio:

\frac{{{3n}\choose{2}}}{3\cdot {{n}\choose{2}}}

you get a lot of canceling, and it simplifies to

\frac{3n-1}{n-1}

which, since spaghetti and hair both have relatively large values of n, is essentially 3.  There’s nothing special about 3 either:  if you split the spaghetti into k pots first, you’d end up with \frac{1}{k} as many potential tangles.

Why isn’t there a square root?  There should be, I thought, because the number of pairs increases like a quadratic as you add more strands.  But as TwoPi pointed out, when I interrupted his reading to discuss this problem, each pot or group of \frac{1}{k} strands has roughly \frac{1}{k^2} as many distinct pairs as the total, but there are also k pots, and that’s where the \frac{1}{k} comes from.  Or, as I thought about it, each individual strand can only tangle with \frac{1}{k} as many strands if you split them up and that carries over into the totals.

So in the end, since I end up with far fewer than 1/3 as many tangles by braiding, it turns out that it has very little to do with the mathematics and pretty much everything to do with the tension in the braid.  I’m afraid that in the great Math-Physics Match, this round goes to Physics.

Isn’t that a great picture from Harper’s Bazaar?  It’s from 1868.   The spaghetti is published under GNU-FDL by Tim ‘Avatar’ Bartel.

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